3.5.8 \(\int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx\) [408]

Optimal. Leaf size=98 \[ -\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

2/3*b*csc(f*x+e)*(b*sec(f*x+e))^(3/2)/f-5/3*b^3*csc(f*x+e)/f/(b*sec(f*x+e))^(1/2)+5/3*b^2*(cos(1/2*f*x+1/2*e)^
2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(b*sec(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.07, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2706, 2705, 3856, 2720} \begin {gather*} -\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2*(b*Sec[e + f*x])^(5/2),x]

[Out]

(-5*b^3*Csc[e + f*x])/(3*f*Sqrt[b*Sec[e + f*x]]) + (5*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*
Sec[e + f*x]])/(3*f) + (2*b*Csc[e + f*x]*(b*Sec[e + f*x])^(3/2))/(3*f)

Rule 2705

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-a)*b*(a*Cs
c[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(m - 1))), x] + Dist[a^2*((m + n - 2)/(m - 1)), Int[(a*Csc[e
+ f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]

Rule 2706

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a*Csc[e
+ f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(n - 1))), x] + Dist[b^2*((m + n - 2)/(n - 1)), Int[(a*Csc[e + f
*x])^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx &=\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac {1}{3} \left (5 b^2\right ) \int \csc ^2(e+f x) \sqrt {b \sec (e+f x)} \, dx\\ &=-\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac {1}{6} \left (5 b^2\right ) \int \sqrt {b \sec (e+f x)} \, dx\\ &=-\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac {1}{6} \left (5 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx\\ &=-\frac {5 b^3 \csc (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 67, normalized size = 0.68 \begin {gather*} \frac {b \left (2-3 \cot ^2(e+f x)+5 \cos ^{\frac {3}{2}}(e+f x) \csc (e+f x) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )\right ) (b \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2*(b*Sec[e + f*x])^(5/2),x]

[Out]

(b*(2 - 3*Cot[e + f*x]^2 + 5*Cos[e + f*x]^(3/2)*Csc[e + f*x]*EllipticF[(e + f*x)/2, 2])*(b*Sec[e + f*x])^(3/2)
*Sin[e + f*x])/(3*f)

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Maple [C] Result contains complex when optimal does not.
time = 0.34, size = 202, normalized size = 2.06

method	result	size
default	\(\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (5 i \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+5 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-5 \left (\cos ^{2}\left (f x +e \right )\right )+2\right ) \cos \left (f x +e \right ) \left (\cos \left (f x +e \right )+1\right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{3 f \sin \left (f x +e \right )^{5}}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/f*(-1+cos(f*x+e))^2*(5*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*cos(f*x+e)^2*sin(f*x+e)+5*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ell
ipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*cos(f*x+e)-5*cos(f*x+e)^2+2)*cos(f*x+e)*(cos(f*x+e)+1)^2*(b/
cos(f*x+e))^(5/2)/sin(f*x+e)^5

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^2, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 143, normalized size = 1.46 \begin {gather*} \frac {-5 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (5 \, b^{2} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{6 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/6*(-5*I*sqrt(2)*b^(5/2)*cos(f*x + e)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))
+ 5*I*sqrt(2)*b^(5/2)*cos(f*x + e)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*
(5*b^2*cos(f*x + e)^2 - 2*b^2)*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)*sin(f*x + e))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**2*(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\sin \left (e+f\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^2,x)

[Out]

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^2, x)

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